2010 Healthcare Law. (6.48, p. 248) On June 28, 2012 the U.S. Supreme Court upheld the much debated 2010 healthcare law, declaring it constitutional. A Gallup poll released the day after this decision indicates that 46% of 1,012 Americans agree with this decision. At a 95% confidence level, this sample has a 3% margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning.

  1. We are 95% confident that between 43% and 49% of Americans in this sample support the decision of the U.S. Supreme Court on the 2010 healthcare law.

Answer:

False. We are 100% confident that 46% of Americans in this sample support the decision of the U.s. Supreme Court on the 2010 healthcare law.
  1. We are 95% confident that between 43% and 49% of Americans support the decision of the U.S. Supreme Court on the 2010 healthcare law.

Answer:

True. As the statement is for all Americans.
  1. If we considered many random samples of 1,012 Americans, and we calculated the sample proportions of those who support the decision of the U.S. Supreme Court, 95% of those sample proportions will be between 43% and 49%.

Answer:

False.  From the fist sample, we are 95% confident that the true population proportion is between 43% and 49%, but it does not imply to sample proportions.
  1. The margin of error at a 90% confidence level would be higher than 3%.

Answer:

False.  Decreasing confidence level will decrease the margin of error.

Legalization of marijuana, Part I. (6.10, p. 216) The 2010 General Social Survey asked 1,259 US res- idents: “Do you think the use of marijuana should be made legal, or not?” 48% of the respondents said it should be made legal.

  1. Is 48% a sample statistic or a population parameter? Explain.

Answer:

It is a sample statistic because the 48% is of the 1259 US residents but not the total population.
  1. Construct a 95% confidence interval for the proportion of US residents who think marijuana should be made legal, and interpret it in the context of the data.

Answer:

The 95% confidence interval is (0.452, 0.508).  We are 95% confident that the true proportion of US residents who think marijuana should be made legal falls within 45.2% to 50.8%.
    p <- 0.48
    n <- 1259
    SE <- sqrt(p*(1-p)/n)
    ME <- 1.96 * SE
    c(p-ME, p+ME)
## [1] 0.4524028 0.5075972
  1. A critic points out that this 95% confidence interval is only accurate if the statistic follows a normal distribution, or if the normal model is a good approximation. Is this true for these data? Explain.

Answer:

As we have more than 10 "successes" and 10 "failures", and the survey is randomly drawn from less than 10% of the US population, by the central limit theorem, we can assume the normal distribution is valid here.
  1. A news piece on this survey’s findings states, “Majority of Americans think marijuana should be legalized.” Based on your confidence interval, is this news piece’s statement justified?

Answer:

The statement is not justified because the 95% confidence interval (45.2%, 50.8%) includes 50% in it, we are not sure if the true proportion is less than or higher than 50%.  It is not suitable to make a claim that "majority" might support legalization.

Legalize Marijuana, Part II. (6.16, p. 216) As discussed in Exercise above, the 2010 General Social Survey reported a sample where about 48% of US residents thought marijuana should be made legal. If we wanted to limit the margin of error of a 95% confidence interval to 2%, about how many Americans would we need to survey ?

Answer:

We need to survey 2398 Americans.
    p <- 0.48
    ME <- 0.02
    n <- p*(1-p)/(ME/1.96)^2
    n
## [1] 2397.158

Sleep deprivation, CA vs. OR, Part I. (6.22, p. 226) According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insuffient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data.

Answer:

\(SE_{\hat{p_{1}} - \hat{p_{2}}} = \sqrt{\frac{p_{1}(1-p_{1})}{n} + \frac{p_{2}(1-p_{2})}{n}}\)

  1. The 95% confidence interval is (-0.0175, 0.00150).

  2. Since the 95% confidence interval contains 0, there is no statistically significant difference between the proportions of Californians and Oregonians who are sleep deprived.

p1 <- 0.08
p2 <- 0.088
n1 <- 11545
n2 <- 4691
PE <- p1-p2
SE <- sqrt((p1*(1-p1))/n1 + (p2*(1-p2))/n2)
ME <- 1.96*SE
c(PE-ME, PE+ME)
## [1] -0.017498128  0.001498128

Barking deer. (6.34, p. 239) Microhabitat factors associated with forage and bed sites of barking deer in Hainan Island, China were examined from 2001 to 2002. In this region woods make up 4.8% of the land, cultivated grass plot makes up 14.7% and deciduous forests makes up 39.6%. Of the 426 sites where the deer forage, 4 were categorized as woods, 16 as cultivated grassplot, and 61 as deciduous forests. The table below summarizes these data.

  1. Write the hypotheses for testing if barking deer prefer to forage in certain habitats over others.

Answer:

H0: Barking deer does not prefer to forage in certain habitats over others.

HA: Barking deer prefer to forage in certain habitats over others.
  1. What type of test can we use to answer this research question?

Answer:

A chi-squared test can be used to answer this research question.
  1. Check if the assumptions and conditions required for this test are satisfied.

Answer:

1. Each case must be independent to all other cases.

2. At least 5 expected cases for each category.  Although we only have 4 observed woods, the expected case for woods would be higher, which meets the condition.
  1. Do these data provide convincing evidence that barking deer prefer to forage in certain habitats over others? Conduct an appropriate hypothesis test to answer this research question.

Answer:

As the p-value is approximately zero, we reject the null hypothesis and accept the alternative hypothesis.
    chisq.test(x=c(4,16,61,345),p=c(0.048,0.147,0.396,0.409))
## 
##  Chi-squared test for given probabilities
## 
## data:  c(4, 16, 61, 345)
## X-squared = 284.06, df = 3, p-value < 2.2e-16

Coffee and Depression. (6.50, p. 248) Researchers conducted a study investigating the relationship between caffeinated coffee consumption and risk of depression in women. They collected data on 50,739 women free of depression symptoms at the start of the study in the year 1996, and these women were followed through 2006. The researchers used questionnaires to collect data on caffeinated coffee consumption, asked each individual about physician-diagnosed depression, and also asked about the use of antidepressants. The table below shows the distribution of incidences of depression by amount of caffeinated coffee consumption.

{

}

  1. What type of test is appropriate for evaluating if there is an association between coffee intake and depression?

Answer:

A chi-squared test is appropriate.
  1. Write the hypotheses for the test you identified in part (a).

Answer:

1. H0: There is no association between coffee intake and depression.

2. HA: There is association between coffee intake and depression.
  1. Calculate the overall proportion of women who do and do not suffer from depression.

Answer:

1. 5.14% women suffer from depression.

2. 94.86% women do not suffer from depression.
    # women who suffer from depression
    2607/50739
## [1] 0.05138059
    # women who do not suffer from depression
    48132/50739
## [1] 0.9486194
  1. Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the test statistic, i.e. (\(Observed - Expected)^2 / Expected\)).

Answer:

1. The expected count for the highlighted cell (373) is 339.98.

2. The contribution of this cell to the test statistic is 3.206.
    exp <- 5.138/100*6617
    exp
## [1] 339.9815
    (373-exp)^2 /exp
## [1] 3.206716
  1. The test statistic is \(\chi^2=20.93\). What is the p-value?

Answer:

The p-value is 0.0003269507.
    chisq <- 20.93
    df <- (5-1)*(2-1)
    p <- 1- pchisq(chisq, df)
    p
## [1] 0.0003269507
  1. What is the conclusion of the hypothesis test?

Answer:

1. The p-value is 0.0003269507 <0.05, so we reject the null hypothesis and accept the alternative hypothesis.

2. There is association between coffee intake and depression.
  1. One of the authors of this study was quoted on the NYTimes as saying it was “too early to recommend that women load up on extra coffee” based on just this study.64 Do you agree with this statement? Explain your reasoning.

Answer:

I agree. The study is from observations but not experiments, which the results may be affected by other factors. It is unclear that coffee intake affects depression in women directly.